Question

Find all forces in Members AB AH BH BC CD CF GF

(a) Determine the forces in members: AB, AH, BH and BC using Method of Joints. (b) Determine the forces in members: CD, CF, and GF using Method of Sections. Н. F 4.5 m 3 m E B DI 2 KN 6 KN 8 KN - 12 m, 4 @ 3 m

Answer 1

G 4.5m H 3m IB ARBE in LORN PRIN GKN 3m 3 m 3m 3m ZFy=o- RA TRE - 6-8-2=0 RA TRE = 16 EME=0 RAX (12) - 649 - 846 - 2a3=0 RA= 9 KN RATRE = 16 9+RE = 16 RE= 7 KN

using method of Joint at at Joint A Fan é 3/45 tano = 3 3 o=tan(1) -45' FAB RA=9kn & Fy=0 FAH sino +RA = 0 Fan sin 45' +9=0 Fan=- 12.728 KN Fan=12.728 KN (Compression) ne & Exco Fotos 4 Fan coso + TAB=0 - 12.728 Cos 45 + FAB = 0 FAB= 9 KN (tensions

at Joint B. и FAB FBC B 6 KN & Fy=0 Fви — бе о Fви – GEN (tension) & fx=0 с – ве = (tension) [Faca Fire - 9 KN ам Au Хиш АВ | -18 te) 9 KN (T) (c) An (2.42 8 (т) ВИ 6 м (Т) БС 9 KN

X 4.5m H 3m D A7KN=RE B LUN 6 KN 8W 9kn 3m 3m 3m using method of section tano 3 4.5-3 453 FGF &= tan " (1) A FLE 3m B ECD $ = 26.565 RA =9kn 6KN & KN 3 m 3m 3m EME=0 Rang - 6x6 - 813 - Flox3=0 gag - 646 – 8x3 - Fox 3=0 .. Fo=7kN (tension)

& Fx=0 For coso t FCF cos o + FLD =0 ForF cos 26.565 t FCF cos 45 +7=0 FGEVO cos 26.365 t Feu cos 45'=-7 -i) E fy=0 FGF sin 26.565=0. -6-8 + FCF. sin 45' P 9-6-8 + FCF sin 45 - FGF sin 26.565=0 FGF in 26.565- FCF bis 45° = – 5 -11) soluery ;) 411) we get FGF=- 8.944 KN FGF = 8.944 KN (compression) FCF = 1.414 KN (tension 7 KN force CD CE GF 11414 KN 8,9 44 KN (teneion) (T) (c)
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